The Golay codes Mario de Boer and Ruud Pellikaan ∗ Appeared in Some tapas of computer algebra (A.M. Cohen, H. Cuypers and H. Sterk eds.), Project 7, The Golay codes, pp. 338-347, Springer, Berlin 1999, after the EIDMA/Galois minicourse on Computer Algebra, September 27-30, 1995, Eindhoven. ∗Both authors are from the Department of Mathematics and Computing Science, Eind- hoven University of Technology, P.O. Box 513, 5600 MB Eindhoven, The Netherlands. 1 Contents 1 Introduction 3 2 Minimal weight codewords of G11 3 3 Decoding of G23 with Gr¨obnerbases 6 4 One-step decoding of G23 7 5 The key equation for G23 9 6 Exercises 10 2 1 Introduction In this project we will give examples of methods described in the previous chap- ters on finding the minimum weight codewords, the decoding of cyclic codes and working with the Mathieu groups. The codes that we use here are the well known Golay codes. These codes are among the most beautiful objects in coding theory, and we would like to give some reasons why. There are two Golay codes: the ternary cyclic code G11 and the binary cyclic code G23. The ternary Golay code G11 has parameters [11, 6, 5], and it is the unique code with these parameters. The automorphism group Aut(G11) is the Mathieu group M11. The group M11 is simple, 4-fold transitive and has size 11 · 10 · 9 · 8. The supports of the codewords of weight 5 form the blocks of a 4-design, the unique Steiner system S(4, 5, 11). The ternary Golay code is a perfect code, this means that the Hamming spheres of radius (d − 1)/2 = 2 centered at the codewords of 11 G11 exactly cover the whole space F3 . The code G11 can be uniquely extended to a [12, 6, 6] code, which we will denote by G12. The code G12 is self-dual and Aut(G12) = M12: the simple, 5-fold transitive Mathieu group of size 12·11·10·9·8. The supports of the codewords of weight 6 in G12 form a 5-design, the unique S(5, 6, 12). The binary Golay code G23 has similar properties. Its parameters are [23, 12, 7], and it is the unique code with these parameters. The automorphism group Aut(G23) is the Mathieu group M23. The group M23 is simple, 4-fold transitive and has size 23 · 22 · 21 · 20 · 48. The supports of the codewords of weight 7 form the blocks of a 4-design, the unique Steiner system S(4, 7, 23). The binary Golay code is a perfect code, so the Hamming spheres of radius 3 centered at 23 the codewords of G11 exactly cover the whole space F2 . The code G23 can be uniquely extended to a [24, 12, 8] code, which we will denote by G24. The code G24 is self-dual and Aut(G24) = M24: the simple, 5-fold transitive Mathieu group of size 24 · 23 · 22 · 21 · 20 · 48. The supports of the codewords of weight 8 in G24 form a 5-design, the unique S(5, 8, 24). 2 Minimal weight codewords of G11 G11 is the ternary cyclic code of length 11 with defining set J = {1}. It is a [11, 6, d] code with complete defining set J(G11) = {1, 3, 4, 5, 9}. The generator polynomial is Y g(X) = (X − αj) = 2 + X2 + 2X3 + X4 + X5. j∈J(G11) From the BCH bound we see that d ≥ 4, and by computing Gr¨obner bases we will show that in fact d = 5. Moreover we will determine all codewords of 3 minimal weight. First we consider the system SG11 (4): A + σ A + σ A + σ A + σ A = 0 5 1 4 2 3 3 2 4 1 A + σ A + σ A + σ A + σ A = 0 6 1 5 2 4 3 3 4 2 . . SG11 (4) = A4 + σ1A3 + σ2A2 + σ3A1 + σ4A0 = 0 Aj = 0 for j ∈ J(G11) 3 A3j = Aj for j = 1,..., 11. 3 Using A3i = Ai we can express every Ai with i ∈ {1, 2,..., 10}\ J(G11) as a power of A2 (this can be done since all of these i form a single cyclotomic coset). Setting Ai = 0 for i ∈ J(G11) and writing A2 = a and A0 = b this reduces SG11 (4) to σ3a = 0 3 a + σ4a = 0 9 3 a + σ1a = 0 81 9 3 a + σ1a + σ2a = 0 81 9 3 σ1a + σ2a + σ3a = 0 27 81 9 3 a + σ2a + σ3a + σ4a = 0 SG11 (4) = 27 81 9 b + σ1a + σ3a + σ4a = 0 27 81 σ1b + σ2a + σ4a = 0 27 a + σ2b + σ3a = 0 27 σ1a + σ3b + σ4a = 0 σ2a + σ4b = 0 b3 − b = 0. Computing a Gr¨obner basis G with respect to the lexicographic order with σ4 > σ3 > σ2 > σ1 > b > a gives G = {b, a} and hence there are no nonzero codewords of weight at most 4. We conclude d ≥ 5, and even d = 5, since the weight of the generator polynomial is wt(g(X)) = 5. To determine the minimum weight codewords we consider the system SG11 (5): A6 + σ1A5 + σ2A4 + σ3A3 + σ4A2 + σ5A1 = 0 A7 + σ1A6 + σ2A5 + σ3A4 + σ4A3 + σ5A2 = 0 . . SG11 (5) = A5 + σ1A4 + σ2A3 + σ3A2 + σ4A1 + σ5A0 = 0 Ai = 0 for i ∈ J(G11) 3 A3i = Ai for i = 0,..., 10 Again we can reduce the system as we did in the system SG11 (4) and compute its Gr¨obner basis with respect to the lexicographic order with σ5 > σ4 > σ3 > σ2 > σ1 > b > a. 4 After 2 minutes using Axiom or 10 minutes using Macaulay, the resulting basis G is 31 9 3 107 41 19 σ5a + 2a + 2a , σ4a + a , σ3a + 2a + a + 2a , 79 35 13 29 7 G = σ2a + a + 2a + a , σ1a + a + 2a , b + a77 + 2a55 + a33 + a11, a133 + 2a111 + 2a89 + 2a67 + a45 + a, where a = A2 and b = A0. From the triangular form of the basis G, it is easy to see that the number of codewords of weight 5 in G11 equals the number of nonzero solutions to f(X) = X133 + 2X111 + 2X89 + 2X67 + X45 + X = 0 in F35 . We determine these solutions in the following exercise. Exercise 2.1 Let α ∈ F35 be a primitive element. Now show 1. f(1) = 0; 2. f(α2) = 0 (you can use a computer algebra package for this); 3. f(α11X) = α11f(X). Conclude from this that the complete set of zeros of f(X) in F35 \{0} is i+11j M = {α | i ∈ {0, 1,..., 10}\ J(G11), j ∈ {0, 1,..., 21}}. So the number of codewords of weight 5 is #M = 132 and the locators of these words (i.e. the polynomials having as zeros the reciprocals of positions where the codewords have a nonzero value) are given by 30 8 5 2 4 (a + a )X + 2a X + (a106 + 2a40 + a18)X3+ σ(X, a) = 78 34 12 2 (2a + a + 2a )X + (2a28 + a6)X + 1, with a ∈ M. Since the code is cyclic, any shift of a codeword of weight 5 is again a codeword of weight 5. We can recognize this fact from M in the following way. Exercise 2.2 Show that there exists a primitive 11-th root of unity β such that σ(X, α11a) = σ(βX, a) for all a ∈ M. Now we can conclude that the codewords of weight 5 consist of the 6 codewords with locator polynomials σ(X, a), a ∈ {1, α2, α6, α7, α8, α10}, their cyclic shifts, 11 and their non-zero multiples in F3 . 22 Exercise 2.3 Let α again be a primitive element in F35 , then β = α is a fixed 11-th root of unity. Check that the zeros of the 6 polynomials σ(X, a) are: polynomial {i | β−i is a zero} σ(X, 1) 2, 6, 7, 8, 10 σ(X, α2) 3, 4, 9, 10, 11 σ(X, α6) 1, 5, 8, 9, 11 σ(X, α7) 1, 2, 8, 10, 11 σ(X, α8) 2, 3, 5, 7, 9 σ(X, α10) 3, 5, 8, 10, 11 5 Let B consists of the 6 subsets in {1,..., 11} of size 5 in the table and their cyclic shifts modulo 11. Then |B| = 66. Show that B is the set of blocks of a 4-design, the Steiner system S(4, 5, 11). 3 Decoding of G23 with Gr¨obnerbases Let G23 be the binary cyclic code of length 23 with defining set J = {1}. Then the complete defining set is J(G23) = {1, 2, 3, 4, 6, 8, 9, 12, 13, 16, 18} and the code has parameters [23, 12, d]. The BCH bound states that d ≥ 5 but in fact d = 7. This can be checked in the same way as we did in the previous section for the ternary Golay code. The computer algebra packages we tried, did not perform very well on the systems SG23 (w). Since the minimum distance is 7, G23 should be able to correct errors of weight at most 3. In this example we will decode a word with three errors. Take 11 2 F211 = F2[β]/(β + β + 1) 89 and set α = β . Then β is a primitive element of F211 and α has order 23. The generator polynomial of the code is Y g(X) = (X − αj) = 1 + X + X5 + X6 + X7 + X9 + X11. j∈J(G23) Suppose we send the codeword g(X), which corresponds to the binary vector c = (1, 1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0) over a noisy channel, and the following error occurs during transmission: e = (1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0).